# An elementary proof of Sobolev's Inequality in one dimension

I was looking for an elementary proof for Sobolev's Inequality in one dimension and I found the proof in this paper, it's the proof of Sobolev imbedding. I would like an explanation about some steps of the proof.

There is a moment when the author of the paper asserts

$|f(y)| + |f(x) - f(y)| \leq \int_0^1 |f(t)|dt + |f(x) - f(y)| \leq \int_0^1 |f| \cdot 1 + |f'|_{L^2} \cdot 1$ $\leq |f|_{L^2} + |f'|_{L^2} << 2 \left( |f|^2 + |f'|^2 \right)^{\frac{1}{2}},$

just a notice that $|f| := \left( \int_a^b |f(t)|^2 dt \right)^{\frac{1}{2}}$

I didn't understand in the first inequality why the author can asserts that $|f(y)| \leq \int_0^1 |f(t)| dt$. For this inequality, I think that we can assume without loss of generality that $f(0) = 0$, because we can translate $f$ so that $f(0) = 0$, therefore we would have

$|f(y)| = |f(y) - f(0)| = | \int_0^y f(t) dt | \leq | \int_0^1 f(t) dt | \leq \int_0^1 |f(t)| dt$

I didn't understand too why $\int_0^1 |f(t)|dt \leq \int_0^1 |f| \cdot 1$.

$\int_0^1 |f| \cdot 1 \leq |f|_{L^2}$

I don't sure, but it seems like that here is used Cauchy-Schwarz inequality.

$|f|_{L^2} + |f'|_{L^2} << 2 \left( |f|^2 + |f'|^2 \right)^{\frac{1}{2}}$

and I don't have idea why he can asserts this inequality.

$\textbf{EDIT1:}$

I forgot to ask why this proof allows to conclude that $\max |f(x)|^2 \leq C \int_0^1 \left( |f|^2 + |f'|^2 \right)$ for some constant $C$.

$\textbf{EDIT2:}$

I correct the EDIT1.

• What is the definition of y? You don’t need to assume f(0)=0. Nov 11, 2017 at 13:44
• @0876, $y$ is just an element of $[0,1]$, i.e, $y \in [0,1]$, sorry for don't put all the details of proof, but the details can be found in the paper that I put in the begin of the topic. Nov 11, 2017 at 14:00
• You should check the definition of y in the paper you mentioned. y is the point where |f| has its minimum. Nov 11, 2017 at 14:02

I will use $\|f\|_X$ for function norms and $|f(x)|$ for absolute values.

You may have missed that the author states that $y$ was such that $|f(y)|= \min_x |f(x)| ≤ |f(z)|$ for every $z\in[0,1]$. Integrating in $z$,

$$|f(y)| = \min_x |f(x)|\int_0^1 dz ≤ \int_0^1|f(z)| \, dz$$ You are correct for the next line, Cauchy Schwartz is used.

I will just write out the proof here,

\begin{align} |f(x)| &\overset{\text{triangle ineq}}≤ |f(y)| + |f(x)-f(y)| \\ &\overset{\text{definition of $y$}}≤ \int_0^1 |f(z)|\,dz + |f(x)-f(y)| \\ &\overset{\text{C-S ineq}}≤ \|f\|_{L^2}\int_0^11^2dz + \|f'\|_{L^2}|y-x|^{1/2} \\ &≤ \|f\|_{L^2} + \|f'\|_{L^2} \\ &≤ C(\|f\|_{L^2}^2 +\|f'\|_{L^2}^2)^{1/2} \end{align}

Where this last line is from the easy inequality $2ab≤a^2+b^2$ so that $$(a+b)^2 ≤ a^2 + b^2 + 2ab ≤ 2a^2+2b^2$$ Take a square root to conclude. This line also follows from the fact that all norms on $\mathbb R^n$ are equivalent.

It should be noted that OP states a wrong inequality $\max_x |f(x)|^2 ≤ C(\|f\|_{L^2}^2 +\|f'\|_{L^2}^2)^{1/2}$ as can be seen from this line of reasoning.

• Firstly, the fact that $2ab \leq a^2 + b^2$ follows that $(a-b)^2 \geq 0$, which implies that $a^2 + b^2 \geq 2ab$, isn't it? Secondly, I can't see yet how this inequality helps to prove the last line, because if we denote $a := ||f||_{L^2}$ and $b := ||f'||_{L^2}$, the last inequality means that $a + b \leq C \sqrt{a^2 + b^2}$, we have here an inequality involving sums and $2ab \leq a^2 + b^2$ is an inequality involving an product and a sum. Nov 11, 2017 at 14:18
• you're so close - the very last line I wrote $(\dots)^2 ≤ \dots$ is the square of the inequality that you want $(\dots) ≤ \sqrt{\dots}$, and I used the inequality of product and sum to prove it. Nov 11, 2017 at 14:25
• @George also, have some faith in yourself, you should be able to verify $(a-b)^2≥0$ proves $2ab ≤ a^2 + b^2$ without me :) Nov 11, 2017 at 14:28
• I verified, I just found strange that you asserts that the inequality $2ab \leq a^2 + b^2$ follows that $(a + b)^2 \geq 0$, because it seems that you use implicitly that $(a - b)^2 \geq 0$. About your previous comment, it's doesn't clear yet for me, but I will think about more about this. Nov 11, 2017 at 14:35
• I do not care about a lower bound for $(a+b)^2$. Moreover I feel that the fact that the square of any real number is postive should be clear to anyone trying to understand sobolev embedding. In any case, the important point is to prove an upper bound that looks like $a^2 + b^2$. Expand $(a+b)^2$ and you see the product $2ab$ naturally appear next to $a^2+b^2$, and use the inequality $2ab ≤ a^2+b^2$. Nov 11, 2017 at 14:40

Looking at the link, I actually don't think the first chain of inequalities is very clearly presented. Here's how I'd say the same thing:

Assume first that $f\in C^\infty$. Now for every $t$ it's clear that $$|f(x)|\le|f(t)|+|f(x)|+|f(x)-f(t)|.$$

Now since $|f'|=|f'|\cdot 1$, Cauchy-Schwarz shows that $$|f(x)-f(t)|=\left|\int_x^tf'\right|\le\int_x^t|f'|\le\int_0^1|f'| \le\left(\int_0^1|f'|^2\right)^{1/2}.$$So we have $$|f(x)|\le|f(t)|+\left(\int_0^1|f'|^2\right)^{1/2}.$$

Now take the integral from $0$ to $1$ of both sides of that last inequality, with respect to $t$. Since $f(x)$ and $\left(\int_0^1|f'|^2\right)^{1/2}$ are independent of $t$ (and of course $\int_0^1c=c$) we obtain $$|f(x)|\le\int_0^1|f(t)|\,dt+\left(\int_0^1|f'|^2\right)^{1/2}.$$Another application of Cauchy-Schwarz shows that $$|f(x)|\le\left(\int_0^1|f|^2\right)^{1/2}+\left(\int_0^1|f'|^2\right)^{1/2}.$$

Now if $|f(x)|\le c$ for every $x$ then $\max_x|f(x)|\le c$. So we've shown that $$\max_x|f(x)|\le \left(\int_0^1|f|^2\right)^{1/2}+\left(\int_0^1|f'|^2\right)^{1/2}.$$

That's the inequality we want, for $f\in C^\infty$. This automatically extends to $f\in H^1$, because of the following fact:

Suppose $X$ and $Y$ are normed vector spaces, $D$ is a dense subspace of $X$, and $T:D\to Y$ is a linear map with $||Tx||\le||x||$ for every $x\in D$. Then $T$ extends to a linear map $T:X\to Y$ with $||Tx||\le||x||$ for every $x\in X$.

(Apply that with $X=H^1$, $D=C^\infty$, and $Y=C([0,1])$, where $X$ has the Sobolev norm, $Y$ has the sup norm, and $Tf=f$.)